A closed surface is defined in spherical coordinates by 3<r<5 , 0.1π<θ<0.3π, 1.2π<φ<1.6π. Find the total surface area.


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Question Number 82591 by Learner-123 last updated on 22/Feb/20

$A c l o s e d s u r f a c e i s d e f i n e d i n s p h e r i c a l$ $c o o r d i n a t e s b y 3 < r < 5, 0.1 π < θ < 0.3 π,$ $1.2 π < ϕ < 1.6 π . F i n d t h e t o t a l s u r f a c e$ $a r e a .$
Commented by Learner-123 last updated on 22/Feb/20

$A n s : 36.79 s q u a r e u n i t s .$
Commented by Learner-123 last updated on 23/Feb/20

Commented by Learner-123 last updated on 23/Feb/20

$S i r, t h i s i s t h e e x p l a i n a t i o n g i v e n .$ $C a n y o u t e l l h o w t h e s e 4 (a c t u a l l y 5) t e r m s$ $c o m e . .$
Commented by mr W last updated on 23/Feb/20

Commented by mr W last updated on 23/Feb/20

$i n o u r c a s e$ $ϕ = 0.1 π . . 0.3 π$ $θ = 1.2 π . . 1.6 π$ $r = 3 . . 5$ $w e h a v e t o t a l l y 6 f a c e s :$ $f a c e 1 : r = 3, ϕ = 0.1 π . . . 0.3 π, θ = 1.2 π . . 1.6 π$ $A_{1} = \int_{1.2 π}^{1.6 π} \int_{0.1 π}^{0.3 π} r \sin θ d ϕ r d θ = 3^{2} \int_{1.2 π}^{1.6 π} \int_{0.1 π}^{0.3 π} \sin θ d ϕ d θ$ $f a c e 2 : r = 5, ϕ = 0.1 π . . . 0.3 π, θ = 1.2 π . . 1.6 π$ $A_{2} = s i m i l a r l y = 5^{2} \int_{1.2 π}^{1.6 π} \int_{0.1 π}^{0.3 π} \sin θ d ϕ d θ$ $f a c e 3 : ϕ = 0.1 π, θ = 1.2 π . . 1.6 π, r = 3 . . 5$ $A_{3} = \frac{1}{2} (5^{2} - 3^{2}) (1.6 π - 1.2 π)$ $f a c e 4 : ϕ = 0.3 π, θ = 1.2 π . . 1.6 π, r = 3 . . 5$ $A_{4} = \frac{1}{2} (5^{2} - 3^{2}) (1.6 π - 1.2 π)$ $f a c e 5 : ϕ = 0.1 π . . 0.3 π, θ = 1.2 π, r = 3 . . 5$ $A_{5} = \int_{0.1 π}^{0.3 π} \int_{3}^{5} \sin 1.2 π d ϕ r d r$ $f a c e 6 : ϕ = 0.1 π . . 0.3 π, θ = 1.6 π, r = 3 . . 5$ $A_{6} = \int_{0.1 π}^{0.3 π} \int_{3}^{5} \sin 1.6 π d ϕ r d r$ $A = Σ A$ $= (3^{2} + 5^{2}) \int_{1.2 π}^{1.6 π} \int_{0.1 π}^{0.3 π} \sin θ d ϕ d θ$ $+ 2 \times \frac{1}{2} (5^{2} - 3^{2}) (1.6 π - 1.2 π)$ $+ (\sin 1.2 π + \sin 1.6 π) (0.3 π - 0.1 π) \int_{3}^{5} r d ϕ d r$
Commented by Learner-123 last updated on 24/Feb/20

$T h a n k Y o u S i r!$ $C a n y o u p l e a s e s o l v e m y o t h e r 3 d o u b t s ?$
Commented by mr W last updated on 24/Feb/20

$i {d o n}^{'} t k n o w w h a t a r e y o u r d o u b t s . y o u$ $s h o u l d n o t a s s u m e t h a t i k n o w a l l p o s t s .$
Commented by Learner-123 last updated on 24/Feb/20

$Q u e s t i o n n o : 82627, 82631, 82647$
Commented by Learner-123 last updated on 24/Feb/20

$I t h i n k t h e r e i s s o m e t e c h n i c a l p r o b l e m$ $i n t h i s a p p, w h e n e v e r i p r e s s ‘ ‘ s o r t b y$ $r e c e n t {a c t i v i t y}^{″} i t g e t s m e t o s a m e$ $i n t e g r a l q u e s t i o n e v e r y t i m e . . . (75986)$
Commented by mr W last updated on 24/Feb/20

$i {c a n}^{'} t h e l p y o u .$
Commented by mr W last updated on 26/Feb/20

$b o t h$
Commented by Learner-123 last updated on 26/Feb/20

$A r e y o u r e f e r r i n g t o m y d o u b t o r$ $t h i s t e c h n i c a l g l i t c h ?$
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